Problem 48 - A Complete Resource Book in Physics for JEE Main

Question

Eight dipoles of charges of magnitude \(e\) are placed inside a cube. The total flux coming out of the cube equals to (A) \(\frac{8 e}{\varepsilon_{0}}\) (B) \(\frac{16 e}{\varepsilon_{0}}\) (C) \(\frac{e}{\varepsilon_{0}}\) (D) Zero

Step-by-Step Solution

Answer

(D) Zero

Step 1: Understand the Nature of a Dipole

Dipoles are pairs of charges of equal magnitude and opposite sign. Therefore, no matter what the magnitude of the individual charges in the dipole, the net charge of a dipole is always zero.

Step 2: Apply Gauss's Law

According to Gauss's Law, the electric flux through a closed surface equals the total net charge enclosed by the surface, divided by the permittivity of free space (\( \varepsilon_{0} \)). Therefore, the total flux coming out of the cube is \( \frac{Net\; Charge\; enclosed}{\varepsilon_{0}} \)

Step 3: Determine the Net Charge Enclosed Within the Cube

Since there are eight dipoles inside the cube, and each dipole's total charge is zero (since the negative and positive charges cancel out), the total charge enclosed inside the cube is also zero.